Chemistry Test - 23

Given, solubility in water \((\mathrm{S})=8.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\)

In pure water

\(\begin{aligned}& \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^2=\left(8 \times 10^{-4}\right)^2=64 \times 10^{-8} \\& \ln 0.01 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4 \\& \mathrm{H}_2 \mathrm{SO}_4(\mathrm{aq}) \rightarrow \underset{0.02}{2 \mathrm{H}^{+}(\mathrm{aq})}+\mathrm{SO}_4^{2-}(\mathrm{aq}) \\& \mathrm{Cd} \mathrm{SO}_4(\mathrm{~s}) \rightleftharpoons \mathrm{Cd}^{2+}(\mathrm{aq})+\mathrm{SO}_4^{2-}{ }^{2-}(\mathrm{aq}) \\& \mathrm{x}_{\mathrm{sp}}=\mathrm{x}(\mathrm{x}+0.01)=64 \times 10^{-8} \\& \mathrm{x}+0.01 \cong 0.01 \mathrm{M} \\& \mathrm{So}, \mathrm{x}(0.01)=64 \times 10^{-8} \\& \mathrm{x}=64 \times 10^{-6} \mathrm{M}\end{aligned}\)

By the below given relation:

\(\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\text { Salt }]}{[\text { Base }]}\)

when \([\) Salt \(]=[\) Base \(]\)

And \(\log 1 = 0\)

So, \(\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}=4.75\)

So, \(\mathrm{pH}=14-4.75=9.25 \)

Given that:

The solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass.

Therefore,

Total amount of solute present in the mixture will be given by,

\(300 \times \frac{25}{100}+400 \times \frac{40}{100}\)

\(=75+160\)

\(=235 \mathrm{~g}\)

Total amount of solution \(=300+400=700 \mathrm{~g}\)

Therefore,

Mass percentage of the solute in the resulting solution\(=\frac{\text{Total amount of solute}}{\text{Total amount of solution}} \times100\)

\(=\frac{235}{700} \times100\)

\(=33.57 \%\)

Then, mass percentage of the solvent in the resulting solution will be:

\(=(100-33.57) \%\)

\(=66.43 \%\)

\(\begin{aligned}& \text { (A) } \mathrm{V}_{\mathrm{e}}=1000 \mathrm{~m} / \mathrm{s} ; \mathrm{h}=6 \times 10^{-34} \mathrm{Js} ; \\& \mathrm{m}_{\mathrm{e}}=9 \times 10^{-31} \mathrm{~kg} \\& \lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6 \times 10^{-34}}{9 \times 10^{-31} \times 1000}=666.67 \times 10^{-9} \mathrm{~m} \\& =666.67 \mathrm{~nm}\end{aligned}\)

(B) The characteristic of electrons emitted is independent of the material of the electrodes of the cathode ray tube.

(c) The cathode rays start from cathode and move towards anode.

(D) The nature of the emitted electrons is independent on the nature of the gas present in cathode ray tube.

The compound having the least molecular mass should contain the minimum amount of sulphur or simply 1 atom of sulphur.

Let the molecular mass of the compound (in amu) be \(x\).

Mass of sulphur in the compond \(=8\%\) of the total molecular mass of \(x\)

\(=\frac{8 x}{100}\)

Since, the molecular mass of one mole of sulphur is 32, the compound must contain this amount of sulphur, i.e.,

\(\frac{8 x}{100}=32\)

which gives:

\(x=400\)

Maximum number of emission lines

\(\begin{aligned} & =\frac{\left(\mathrm{n}_2-\mathrm{n}_1\right)\left(\mathrm{n}_2-\mathrm{n}_1+1\right)}{2} \\ & \mathrm{n}_2=5 \\ & \mathrm{n}_1=1 \\ & \Rightarrow \frac{(5-1)(5-1+1)}{2}=10\end{aligned}\)
Hence maximum number of emission lines observed are 10

Hybridization of \(\mathrm{Fe}\) in \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) is \(d^{2} s p^{3}\).

In the given complex of \(\mathrm{Fe}^{3+}\), as the co-ordination number is ' 6 ', there are two possibilities. It can be \(d^{2} s p^{3}\) or \(s p^{3} d^{2}\). But the ligand attached with central metal ion, \(\mathrm{Fe}^{3+}\), is strong field ligand so it will promote inner orbital complex, \(d^{2} s p^{3}\).

Given,

Wavelength of violet light \(=400 \mathrm{~nm} =400 \times 10^{-9} \mathrm{~m} \)

Wavelength of red light \(=750 \mathrm{~nm}=750 \times 10^{-9} \mathrm{~m}\)

As we know,

The speed of light,

\(c= 3 \times 10^8\) m/s

Frequency of violetlight is given as,

\(\nu=\frac{c}{\lambda}\)

\(=\frac{3.00 \times 10^{8}}{400 \times 10^{-9}}\)

\(=7.50 \times 10^{14}\) Hz

Frequency of red light is given as,

\(\nu=\frac{c}{\lambda}\)

\(=\frac{3.00 \times 10^{8}}{750 \times 10^{-9}}\)

\(=4 \times 10^{14}\) Hz

The simplification strategy of drug design is frequently used on complex lead compounds derived from natural products.

Structural simplification is a powerful strategy for improving the efficiency and success rate of drug design. For large or complex lead compounds, structural simplification is helpful to discover drug-like molecules with improved synthetic accessibility and favorable pharmacodynamic/pharmacokinetic profiles.

If the amino group of glycine and carboxylic acid group of alanine undergo elimination of water molecule, the formed compound is glycylalanine (dipeptide).